In the ten problems listed below, your task is to make the number shown in bold, using once and once only, all six numbers, which accompany it.
You may use any mathematical operations you wish, to arrive at the given number.

            784 1, 1, 5, 6, 8, 100
            327 6, 7, 8, 9, 9, 50
            931 3, 4, 7, 8, 10, 75
            425 2, 4, 6, 8, 9, 50
            489 2, 3, 4, 6, 10, 75
            845 4, 7, 8, 9, 9, 25
            763 2, 3, 4, 5, 6, 25
            599 2, 3, 4, 6, 7, 75
            291 4, 8, 9, 9, 10, 100
            143 1, 4, 5, 6, 9, 10

Is more than one solution possible for a problem? Please include them if you find some.

There is definitely more than one solution for each. Here are just a few for 784 using 1,1,5,6,8,100:

784 = ((6 * 100) + 5!) + 8^{(1 + 1)}
    = (600 + 120) + 8²
    = 720 + 64

784 = ((!(5 - 1) - (6 - 1)) * 100) + 8!!
    = ((!4 - 5) * 100) + 384
    = ((9 - 5) * 100) + 384
    = (4 * 100) + 384
    = 400 + 384

784 = (6! + (100/1)) - (!5 - (8/1))
    = (720 + 100) - (44 - 8)
    = 820 - 36

784 = (8 * 100) - (!(5 - 1) + (6 + 1))
    = 800 - (!4 + 7)
    = 800 - (9 + 7)
    = 800 - 16

784 = (8/1) * (100 + 5 - 6 - 1)
    = 8 * 98
   

Here are just a few for 327 using 6,7,8,9,9,50:

327 = (8!! + 9) - ((50 + 6 + 7) + SQRT(9))
    = (384 + 9) - (63 + 3)
    = 393 - 66

327 = (50 * 6) + (9 + 8 + 7) + SQRT(9)
    = 300 + 24 + 3

327 = !6 + 50 + (9 + 9 - 8 - 7)$
    = 265 + 50 + 3$
    = 315 + 12

327 = (!6 + (50 + 9 + 8 - 7)) + !(SQRT(9))
    = (265 + 60) + !3
    = 325 + 2

Edited on November 1, 2006, 3:30 pm

Posted by Dej Mar on 2006-11-01 04:29:28